Existence of a group $G\subset O(n)$ along with a homomorphism $\phi :G \to \mathbb{Z}_2 = \{-1,1\}$ with some properties

Question

For studying symmetries of certain PDEs, it would be convenient if a certain type of group existed.

I am looking for a closed subgroup $G$ of the the orthogonal group $O(n)$ along with a continuous homomorphism $\phi : G\to \mathbb{Z}_2 = \{-1, 1\}$ such that

1. If $x\in\mathbb{R}^n,\, x\ne 0$ then $\operatorname{dim}(Gx) > 0$.
2. $\phi : G\to \mathbb{Z}_2$ is surjective.
3. There exists $x\in \mathbb{R^n}$ such that $\{g\in G : gx = x\}\subseteq \operatorname{ker}\phi$.

I get the feeling such a group does not exist. As I am not great with groups/geometry, I was hoping someone could help lead me in the right direction (or direct me towards appropriate references).

Answer 1

$\DeclareMathOperator\SO{SO}$Fix even $n=2m\ge 4$. Let $G_0$ be the copy of $\SO(m)\times\SO(m)$ in $\SO(2m)$. Let $G$ be generated by $G_0$ and an element of order 2 switching the two copies of $\mathbf{R}^m$. Let $x$ be a vector whose decomposition $x_1+x_2$ according to the orthogonal decomposition $\mathbf{R}^m\oplus\mathbf{R}^m$ satisfies $0<\|x_1\|<\|x_2\|$. Then $G$ satisfies all conditions: each orbit has dimension $\ge m-1$, $\phi$ (the homomorphism with kernel $G_0$) is surjective, and $x$ has stabilizer contained in $G_0=\operatorname{ker}\phi$.